类型不可分配给类型 2322

Question

为什么会引发以下问题：Type '(...args: T) => void' is not assignable to type 'Fn<T>'.(2322)：

type Fn<T extends unknown[]|unknown> = T extends unknown[] ? (...args: T)=>void : T

const func = <T extends unknown[]>()=>{
    // why does this error and how does one resolve it?
    // ideal Fn must handle cases outside of unknown[] including `any` and `unknown`
    const fn : Fn<T> = (...args: T)=> console.log(args)
    return fn
}

// some tests by me to try and understand what is going on:
type ValidFn1 = Fn<[value:number]>
type ValidFn2 = Fn<any>
type ValidFn3 = Fn<never>
type ValidFn4 = Fn<unknown>
type ValidFn5 = Fn<unknown[]>
type ValidFn6 = Fn<void>

const validFn1: ValidFn1 = (...args)=>console.log(args) // extends unknown[]
const validFn2: ValidFn2 = (...args:any)=>console.log(args) // any extends unknown[]
const validFn3: ValidFn3 = (...args:never)=>console.log(args) // never doesn't extend unknown[]
const validFn4: ValidFn4 = (...args:unknown)=>console.log(args) // unknown doesn't extend unknown[]
const validFn5: ValidFn5 = (...args:unknown[])=>console.log(args) // extends unknown[]
const validFn6: ValidFn6 = (...args:void)=>console.log(args) // void doesn't extend unknown[]

code

Answer 1

我相信这里发生的事情是以下事实的结合：(1) Typescript 在提供具体实例化之前不会解析函数内部的条件类型（请参阅 ) and (2) function parameters are contravariant (see ）。

所以，你在这里看到的基本上是 TS 看到这一行：

    const fn : Fn<T> = (...args: T)=> console.log(args)

并且：

将 func 中的 Fn<T> 解析为 Fn<unknown[]> ，即 (...args: unknown[]) => void，（因为上面的第 1 点）
意识到右侧的 T 可能比 unknown[] 更窄（例如，number[]）。

然后它认为，嗯，这可能行不通，因为，这可能是说，例如：

   const fn:(args:unknown[])=>void = (args:number[])=>console.log(args)

这是一个错误，因为 args 是逆变的（参见上面的第 2 点）。

你是怎么解决的？我认为解决第 1 点类别下问题的最简单方法是在函数上使用显式 return 类型，然后仅在函数本身内部进行断言，例如：

const func = <T extends unknown[]>():Fn<T>=>{
    const fn = (...args: T)=> console.log(args)
    return fn as Fn<T>
}

类型不可分配给类型 2322

Type is not assignable to type 2322

typescript

typescript-typings

typescript-generics